Inductance in Kirchoff's voltage law

[naftali]

Hi,

when summing the votages when using Kirchoff's voltage law around some loop, why do anywhere they take L(dI/dt) as the voltage fall on the inductance (if we sum the voltage in the current direction)?
If I understand well , if there is a current in some direction, the induced voltage (according to Lenz's law) will be in the opposite direction, so it will tend to cancel the original current, so the voltage will be -L(dI/dt).

Am I wrong?

 

[cabraham]

It depends on whether the inductor is receiving energy, or delivering it.

Claude

 

[naftali]

It depends on whether the inductor is receiving energy, or delivering it.

Can you explain? suppose the inductor is a solenoid, when does it recives or delivers energy?
(I'm talking about DC)

 

[uart]

People often like to get pedantic about including a minus sign in all equations related to induced voltage, so as to reflect Lenz's Law, but in reality the minus sign is entirely meaningless unless a reference direction for the voltage is also given. Include a minus sign if you wish but then you must also change the reference direction for that voltage so that it's the opposite of what we use for resistors. That would be confusing though, so better not to bother.

In fact we also write V = RI for a resistor when there is every bit as stronger case for the minus sign here (as in V = -RI ). After all, the voltage drop accoss a resistor is certainly always in such a direction so as to oppose the current flow in the external circuit! This is not really any different to Lenz's law except that for resistors it's the current that's opposed directly rather than it's rate of change.

 

[naftali]

In fact we also write V = RI for a resistor when there is every bit as stronger case for the minus sign here (as in V = -RI ). After all, the voltage drop accoss a resistor is certainly always in such a direction so as to oppose the current flow in the external circuit!

I don't understand why, if I understand well, the voltage drop is positive when going from positive to negative (relative) voltage, and that is the direction of the current too.
Isn't that right?

 

[uart]

I don't understand why, if I understand well, the voltage drop is positive when going from positive to negative (relative) voltage, and that is the direction of the current too.
Isn't that right?

Yes and that is precisely the direction of voltage that opposes the flow of current in the external circuit. So this really is exactly the same as for an inductor as per Lenz's law, except you'd replace "opposes the flow of current" for the resistor with "opposes the change in current" for an inductor.

The bottom line is this. If you were to write V = -L di/dt then you'd also have to write V = -RI or else you'd end up with two inconsistent reference directions for the voltage of resistors and inductors.

As an analogy (if you're familiar with mechanical systems) we might well insist that the formula for frictional force is never written as F = u N but always as F = -u N to reflect the fact that frictional forces always oppose the direction of motion. There is absolutely nothing wrong with doing this, it's just you'd have to show the frictional force (on your free body diagram for example) as being in the direction of motion and then ascribe a negative value to that force. Most people would just prefer that the force is drawn in the direction that it actually acts and has a positive value.

 

[naftali]

First, thanks for the answers.

Yes and that direction of voltage opposes the flow of current in the external circuit.

I think that I miss something basic.
I attach a simple circuit, the direction of the curret in the external circuit is marked by the arrow. I can't understand : why will the current in the resistor be in the opposite direction?

 

[uart]

I can't yet see your attachment naftali, but let me tell you what you're missing. It's the same mistake that very many people make when trying to apply Lenz's law. The direction of the induced voltage is in a direction which opposes the change of current in the external circuit. Now that's the third time I've said it and the second time I've highlighted it (italics before and now bold). I can not overstate the importance of understanding this aspect.

Don't look at the direction of that voltage with respect to the inductor (or resistor) internally but instead look at it in relation to the external circuit (the rest of the circuit). This is how Lenz's law is supposed to be applied and it's a very important thing to understand.

 

[Bob S]

Hi,

when summing the votages when using Kirchoff's voltage law around some loop, why do anywhere they take L(dI/dt) as the voltage fall on the inductance (if we sum the voltage in the current direction)?
If I understand well , if there is a current in some direction, the induced voltage (according to Lenz's law) will be in the opposite direction, so it will tend to cancel the original current, so the voltage will be -L(dI/dt).

Am I wrong?

If you have an inductance and resistance in series, and a current I₀ sin(wt), then the total voltage drop is
V(wt) = I₀R sin(wt) + I₀ L (d/dt) sin(wt)
V(wt) = I₀R sin(wt) + I₀ L w cos(wt)
V(w) = I(w) R + j w L I(w)

where w = 2 pi f and j means 90 degree phase shift.

Bob S